Paper 2, Section II, A

Methods | Part IB, 2016

Consider a bar of length π\pi with free ends, subject to longitudinal vibrations. You may assume that the longitudinal displacement y(x,t)y(x, t) of the bar satisfies the wave equation with some wave speed cc :

2yt2=c22yx2\frac{\partial^{2} y}{\partial t^{2}}=c^{2} \frac{\partial^{2} y}{\partial x^{2}}

for x(0,π)x \in(0, \pi) and t>0t>0 with boundary conditions:

yx(0,t)=yx(π,t)=0\frac{\partial y}{\partial x}(0, t)=\frac{\partial y}{\partial x}(\pi, t)=0

for t>0t>0. The bar is initially at rest so that

yt(x,0)=0\frac{\partial y}{\partial t}(x, 0)=0

for x(0,π)x \in(0, \pi), with a spatially varying initial longitudinal displacement given by

y(x,0)=bxy(x, 0)=b x

for x(0,π)x \in(0, \pi), where bb is a real constant.

(a) Using separation of variables, show that

y(x,t)=bπ24bπn=1cos[(2n1)x]cos[(2n1)ct](2n1)2y(x, t)=\frac{b \pi}{2}-\frac{4 b}{\pi} \sum_{n=1}^{\infty} \frac{\cos [(2 n-1) x] \cos [(2 n-1) c t]}{(2 n-1)^{2}}

(b) Determine a periodic function P(x)P(x) such that this solution may be expressed as

y(x,t)=12[P(x+ct)+P(xct)]y(x, t)=\frac{1}{2}[P(x+c t)+P(x-c t)]

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